\(\int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx\) [135]

Optimal result

Integrand size = 24, antiderivative size = 202 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=-\frac {a \left (48 b^2 c^2-5 a d (16 b c-7 a d)\right ) \sqrt {a x+b x^2}}{64 b^4}+\frac {\left (48 b^2 c^2-80 a b c d+35 a^2 d^2\right ) x \sqrt {a x+b x^2}}{96 b^3}+\frac {d (16 b c-7 a d) x^2 \sqrt {a x+b x^2}}{24 b^2}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}+\frac {a^2 \left (48 b^2 c^2-5 a d (16 b c-7 a d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{64 b^{9/2}} \] Output:

-1/64*a*(48*b^2*c^2-5*a*d*(-7*a*d+16*b*c))*(b*x^2+a*x)^(1/2)/b^4+1/96*(35* 
a^2*d^2-80*a*b*c*d+48*b^2*c^2)*x*(b*x^2+a*x)^(1/2)/b^3+1/24*d*(-7*a*d+16*b 
*c)*x^2*(b*x^2+a*x)^(1/2)/b^2+1/4*d^2*x^3*(b*x^2+a*x)^(1/2)/b+1/64*a^2*(48 
*b^2*c^2-5*a*d*(-7*a*d+16*b*c))*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/b^(9/ 
2)
 

Mathematica [A] (verified)

Time = 0.78 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=\frac {\sqrt {x} \left (\sqrt {b} \sqrt {x} (a+b x) \left (-105 a^3 d^2+10 a^2 b d (24 c+7 d x)+16 b^3 x \left (6 c^2+8 c d x+3 d^2 x^2\right )-8 a b^2 \left (18 c^2+20 c d x+7 d^2 x^2\right )\right )+6 a^2 \left (48 b^2 c^2-80 a b c d+35 a^2 d^2\right ) \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{-\sqrt {a}+\sqrt {a+b x}}\right )\right )}{192 b^{9/2} \sqrt {x (a+b x)}} \] Input:

Integrate[(x^2*(c + d*x)^2)/Sqrt[a*x + b*x^2],x]
 

Output:

(Sqrt[x]*(Sqrt[b]*Sqrt[x]*(a + b*x)*(-105*a^3*d^2 + 10*a^2*b*d*(24*c + 7*d 
*x) + 16*b^3*x*(6*c^2 + 8*c*d*x + 3*d^2*x^2) - 8*a*b^2*(18*c^2 + 20*c*d*x 
+ 7*d^2*x^2)) + 6*a^2*(48*b^2*c^2 - 80*a*b*c*d + 35*a^2*d^2)*Sqrt[a + b*x] 
*ArcTanh[(Sqrt[b]*Sqrt[x])/(-Sqrt[a] + Sqrt[a + b*x])]))/(192*b^(9/2)*Sqrt 
[x*(a + b*x)])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1262, 27, 1221, 1134, 1160, 1091, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx\)

\(\Big \downarrow \) 1262

\(\displaystyle \frac {\int \frac {x^2 \left (8 b c^2+d (16 b c-7 a d) x\right )}{2 \sqrt {b x^2+a x}}dx}{4 b}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {x^2 \left (8 b c^2+d (16 b c-7 a d) x\right )}{\sqrt {b x^2+a x}}dx}{8 b}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1221

\(\displaystyle \frac {\frac {\left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \int \frac {x^2}{\sqrt {b x^2+a x}}dx}{6 b}+\frac {d x^2 \sqrt {a x+b x^2} (16 b c-7 a d)}{3 b}}{8 b}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1134

\(\displaystyle \frac {\frac {\left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \int \frac {x}{\sqrt {b x^2+a x}}dx}{4 b}\right )}{6 b}+\frac {d x^2 \sqrt {a x+b x^2} (16 b c-7 a d)}{3 b}}{8 b}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1160

\(\displaystyle \frac {\frac {\left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x+b x^2}}{b}-\frac {a \int \frac {1}{\sqrt {b x^2+a x}}dx}{2 b}\right )}{4 b}\right )}{6 b}+\frac {d x^2 \sqrt {a x+b x^2} (16 b c-7 a d)}{3 b}}{8 b}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {\frac {\left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x+b x^2}}{b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}}{b}\right )}{4 b}\right )}{6 b}+\frac {d x^2 \sqrt {a x+b x^2} (16 b c-7 a d)}{3 b}}{8 b}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\left (\frac {x \sqrt {a x+b x^2}}{2 b}-\frac {3 a \left (\frac {\sqrt {a x+b x^2}}{b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{3/2}}\right )}{4 b}\right ) \left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right )}{6 b}+\frac {d x^2 \sqrt {a x+b x^2} (16 b c-7 a d)}{3 b}}{8 b}+\frac {d^2 x^3 \sqrt {a x+b x^2}}{4 b}\)

Input:

Int[(x^2*(c + d*x)^2)/Sqrt[a*x + b*x^2],x]
 

Output:

(d^2*x^3*Sqrt[a*x + b*x^2])/(4*b) + ((d*(16*b*c - 7*a*d)*x^2*Sqrt[a*x + b* 
x^2])/(3*b) + ((48*b^2*c^2 - 80*a*b*c*d + 35*a^2*d^2)*((x*Sqrt[a*x + b*x^2 
])/(2*b) - (3*a*(Sqrt[a*x + b*x^2]/b - (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b 
*x^2]])/b^(3/2)))/(4*b)))/(6*b))/(8*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1134

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 
 1))), x] + Simp[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1)))   Int[(d + e*x)^ 
(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[ 
c*d^2 - b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2 
*p]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] 
 && NeQ[m + 2*p + 2, 0]
 

rule 1262

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + b 
*x + c*x^2)^(p + 1)/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m 
 + n + 2*p + 1))   Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^n*(m 
 + n + 2*p + 1)*(f + g*x)^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n + e*g^n*( 
m + p + n)*(d + e*x)^(n - 2)*(b*d - 2*a*e + (2*c*d - b*e)*x), x], x], x] /; 
 FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && 
IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(\frac {\frac {35 a^{2} \left (a^{2} d^{2}-\frac {16}{7} a b c d +\frac {48}{35} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )}{64}-\frac {35 \left (\frac {48 \left (\frac {7}{18} d^{2} x^{2}+\frac {10}{9} c d x +c^{2}\right ) a \,b^{\frac {5}{2}}}{35}-\frac {32 \left (\frac {1}{2} d^{2} x^{2}+\frac {4}{3} c d x +c^{2}\right ) x \,b^{\frac {7}{2}}}{35}+d \,a^{2} \left (\left (-\frac {2 d x}{3}-\frac {16 c}{7}\right ) b^{\frac {3}{2}}+\sqrt {b}\, a d \right )\right ) \sqrt {x \left (b x +a \right )}}{64}}{b^{\frac {9}{2}}}\)	\(132\)
risch	\(-\frac {\left (-48 b^{3} d^{2} x^{3}+56 a \,b^{2} d^{2} x^{2}-128 b^{3} c d \,x^{2}-70 a^{2} b \,d^{2} x +160 a \,b^{2} c d x -96 b^{3} c^{2} x +105 a^{3} d^{2}-240 a^{2} b c d +144 a \,c^{2} b^{2}\right ) x \left (b x +a \right )}{192 b^{4} \sqrt {x \left (b x +a \right )}}+\frac {a^{2} \left (35 a^{2} d^{2}-80 a b c d +48 b^{2} c^{2}\right ) \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{128 b^{\frac {9}{2}}}\)	\(164\)
default	\(c^{2} \left (\frac {x \sqrt {b \,x^{2}+a x}}{2 b}-\frac {3 a \left (\frac {\sqrt {b \,x^{2}+a x}}{b}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+d^{2} \left (\frac {x^{3} \sqrt {b \,x^{2}+a x}}{4 b}-\frac {7 a \left (\frac {x^{2} \sqrt {b \,x^{2}+a x}}{3 b}-\frac {5 a \left (\frac {x \sqrt {b \,x^{2}+a x}}{2 b}-\frac {3 a \left (\frac {\sqrt {b \,x^{2}+a x}}{b}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )}{8 b}\right )+2 c d \left (\frac {x^{2} \sqrt {b \,x^{2}+a x}}{3 b}-\frac {5 a \left (\frac {x \sqrt {b \,x^{2}+a x}}{2 b}-\frac {3 a \left (\frac {\sqrt {b \,x^{2}+a x}}{b}-\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )\)	\(302\)

Input:

int(x^2*(d*x+c)^2/(b*x^2+a*x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

35/64*(a^2*(a^2*d^2-16/7*a*b*c*d+48/35*b^2*c^2)*arctanh((x*(b*x+a))^(1/2)/ 
x/b^(1/2))-(48/35*(7/18*d^2*x^2+10/9*c*d*x+c^2)*a*b^(5/2)-32/35*(1/2*d^2*x 
^2+4/3*c*d*x+c^2)*x*b^(7/2)+d*a^2*((-2/3*d*x-16/7*c)*b^(3/2)+b^(1/2)*a*d)) 
*(x*(b*x+a))^(1/2))/b^(9/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.69 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=\left [\frac {3 \, {\left (48 \, a^{2} b^{2} c^{2} - 80 \, a^{3} b c d + 35 \, a^{4} d^{2}\right )} \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (48 \, b^{4} d^{2} x^{3} - 144 \, a b^{3} c^{2} + 240 \, a^{2} b^{2} c d - 105 \, a^{3} b d^{2} + 8 \, {\left (16 \, b^{4} c d - 7 \, a b^{3} d^{2}\right )} x^{2} + 2 \, {\left (48 \, b^{4} c^{2} - 80 \, a b^{3} c d + 35 \, a^{2} b^{2} d^{2}\right )} x\right )} \sqrt {b x^{2} + a x}}{384 \, b^{5}}, -\frac {3 \, {\left (48 \, a^{2} b^{2} c^{2} - 80 \, a^{3} b c d + 35 \, a^{4} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - {\left (48 \, b^{4} d^{2} x^{3} - 144 \, a b^{3} c^{2} + 240 \, a^{2} b^{2} c d - 105 \, a^{3} b d^{2} + 8 \, {\left (16 \, b^{4} c d - 7 \, a b^{3} d^{2}\right )} x^{2} + 2 \, {\left (48 \, b^{4} c^{2} - 80 \, a b^{3} c d + 35 \, a^{2} b^{2} d^{2}\right )} x\right )} \sqrt {b x^{2} + a x}}{192 \, b^{5}}\right ] \] Input:

integrate(x^2*(d*x+c)^2/(b*x^2+a*x)^(1/2),x, algorithm="fricas")
 

Output:

[1/384*(3*(48*a^2*b^2*c^2 - 80*a^3*b*c*d + 35*a^4*d^2)*sqrt(b)*log(2*b*x + 
 a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + 2*(48*b^4*d^2*x^3 - 144*a*b^3*c^2 + 24 
0*a^2*b^2*c*d - 105*a^3*b*d^2 + 8*(16*b^4*c*d - 7*a*b^3*d^2)*x^2 + 2*(48*b 
^4*c^2 - 80*a*b^3*c*d + 35*a^2*b^2*d^2)*x)*sqrt(b*x^2 + a*x))/b^5, -1/192* 
(3*(48*a^2*b^2*c^2 - 80*a^3*b*c*d + 35*a^4*d^2)*sqrt(-b)*arctan(sqrt(b*x^2 
 + a*x)*sqrt(-b)/(b*x + a)) - (48*b^4*d^2*x^3 - 144*a*b^3*c^2 + 240*a^2*b^ 
2*c*d - 105*a^3*b*d^2 + 8*(16*b^4*c*d - 7*a*b^3*d^2)*x^2 + 2*(48*b^4*c^2 - 
 80*a*b^3*c*d + 35*a^2*b^2*d^2)*x)*sqrt(b*x^2 + a*x))/b^5]
 

Sympy [A] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.40 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=\begin {cases} \frac {3 a^{2} \left (- \frac {5 a \left (- \frac {7 a d^{2}}{8 b} + 2 c d\right )}{6 b} + c^{2}\right ) \left (\begin {cases} \frac {\log {\left (a + 2 \sqrt {b} \sqrt {a x + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: \frac {a^{2}}{b} \neq 0 \\\frac {\left (\frac {a}{2 b} + x\right ) \log {\left (\frac {a}{2 b} + x \right )}}{\sqrt {b \left (\frac {a}{2 b} + x\right )^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a x + b x^{2}} \left (- \frac {3 a \left (- \frac {5 a \left (- \frac {7 a d^{2}}{8 b} + 2 c d\right )}{6 b} + c^{2}\right )}{4 b^{2}} + \frac {d^{2} x^{3}}{4 b} + \frac {x^{2} \left (- \frac {7 a d^{2}}{8 b} + 2 c d\right )}{3 b} + \frac {x \left (- \frac {5 a \left (- \frac {7 a d^{2}}{8 b} + 2 c d\right )}{6 b} + c^{2}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {2 \left (\frac {c^{2} \left (a x\right )^{\frac {5}{2}}}{5} + \frac {2 c d \left (a x\right )^{\frac {7}{2}}}{7 a} + \frac {d^{2} \left (a x\right )^{\frac {9}{2}}}{9 a^{2}}\right )}{a^{3}} & \text {for}\: a \neq 0 \\\tilde {\infty } \left (\frac {c^{2} x^{3}}{3} + \frac {c d x^{4}}{2} + \frac {d^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \] Input:

integrate(x**2*(d*x+c)**2/(b*x**2+a*x)**(1/2),x)
 

Output:

Piecewise((3*a**2*(-5*a*(-7*a*d**2/(8*b) + 2*c*d)/(6*b) + c**2)*Piecewise( 
(log(a + 2*sqrt(b)*sqrt(a*x + b*x**2) + 2*b*x)/sqrt(b), Ne(a**2/b, 0)), (( 
a/(2*b) + x)*log(a/(2*b) + x)/sqrt(b*(a/(2*b) + x)**2), True))/(8*b**2) + 
sqrt(a*x + b*x**2)*(-3*a*(-5*a*(-7*a*d**2/(8*b) + 2*c*d)/(6*b) + c**2)/(4* 
b**2) + d**2*x**3/(4*b) + x**2*(-7*a*d**2/(8*b) + 2*c*d)/(3*b) + x*(-5*a*( 
-7*a*d**2/(8*b) + 2*c*d)/(6*b) + c**2)/(2*b)), Ne(b, 0)), (2*(c**2*(a*x)** 
(5/2)/5 + 2*c*d*(a*x)**(7/2)/(7*a) + d**2*(a*x)**(9/2)/(9*a**2))/a**3, Ne( 
a, 0)), (zoo*(c**2*x**3/3 + c*d*x**4/2 + d**2*x**5/5), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.46 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a x} d^{2} x^{3}}{4 \, b} + \frac {2 \, \sqrt {b x^{2} + a x} c d x^{2}}{3 \, b} - \frac {7 \, \sqrt {b x^{2} + a x} a d^{2} x^{2}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a x} c^{2} x}{2 \, b} - \frac {5 \, \sqrt {b x^{2} + a x} a c d x}{6 \, b^{2}} + \frac {35 \, \sqrt {b x^{2} + a x} a^{2} d^{2} x}{96 \, b^{3}} + \frac {3 \, a^{2} c^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {5}{2}}} - \frac {5 \, a^{3} c d \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {7}{2}}} + \frac {35 \, a^{4} d^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {9}{2}}} - \frac {3 \, \sqrt {b x^{2} + a x} a c^{2}}{4 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a x} a^{2} c d}{4 \, b^{3}} - \frac {35 \, \sqrt {b x^{2} + a x} a^{3} d^{2}}{64 \, b^{4}} \] Input:

integrate(x^2*(d*x+c)^2/(b*x^2+a*x)^(1/2),x, algorithm="maxima")
 

Output:

1/4*sqrt(b*x^2 + a*x)*d^2*x^3/b + 2/3*sqrt(b*x^2 + a*x)*c*d*x^2/b - 7/24*s 
qrt(b*x^2 + a*x)*a*d^2*x^2/b^2 + 1/2*sqrt(b*x^2 + a*x)*c^2*x/b - 5/6*sqrt( 
b*x^2 + a*x)*a*c*d*x/b^2 + 35/96*sqrt(b*x^2 + a*x)*a^2*d^2*x/b^3 + 3/8*a^2 
*c^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 5/8*a^3*c*d*lo 
g(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 35/128*a^4*d^2*log(2* 
b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(9/2) - 3/4*sqrt(b*x^2 + a*x)*a*c 
^2/b^2 + 5/4*sqrt(b*x^2 + a*x)*a^2*c*d/b^3 - 35/64*sqrt(b*x^2 + a*x)*a^3*d 
^2/b^4
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=\frac {1}{192} \, \sqrt {b x^{2} + a x} {\left (2 \, {\left (4 \, {\left (\frac {6 \, d^{2} x}{b} + \frac {16 \, b^{3} c d - 7 \, a b^{2} d^{2}}{b^{4}}\right )} x + \frac {48 \, b^{3} c^{2} - 80 \, a b^{2} c d + 35 \, a^{2} b d^{2}}{b^{4}}\right )} x - \frac {3 \, {\left (48 \, a b^{2} c^{2} - 80 \, a^{2} b c d + 35 \, a^{3} d^{2}\right )}}{b^{4}}\right )} - \frac {{\left (48 \, a^{2} b^{2} c^{2} - 80 \, a^{3} b c d + 35 \, a^{4} d^{2}\right )} \log \left ({\left | 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a \right |}\right )}{128 \, b^{\frac {9}{2}}} \] Input:

integrate(x^2*(d*x+c)^2/(b*x^2+a*x)^(1/2),x, algorithm="giac")
 

Output:

1/192*sqrt(b*x^2 + a*x)*(2*(4*(6*d^2*x/b + (16*b^3*c*d - 7*a*b^2*d^2)/b^4) 
*x + (48*b^3*c^2 - 80*a*b^2*c*d + 35*a^2*b*d^2)/b^4)*x - 3*(48*a*b^2*c^2 - 
 80*a^2*b*c*d + 35*a^3*d^2)/b^4) - 1/128*(48*a^2*b^2*c^2 - 80*a^3*b*c*d + 
35*a^4*d^2)*log(abs(2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) + a))/b^(9/2 
)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=\int \frac {x^2\,{\left (c+d\,x\right )}^2}{\sqrt {b\,x^2+a\,x}} \,d x \] Input:

int((x^2*(c + d*x)^2)/(a*x + b*x^2)^(1/2),x)
 

Output:

int((x^2*(c + d*x)^2)/(a*x + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.27 \[ \int \frac {x^2 (c+d x)^2}{\sqrt {a x+b x^2}} \, dx=\frac {-105 \sqrt {x}\, \sqrt {b x +a}\, a^{3} b \,d^{2}+240 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} c d +70 \sqrt {x}\, \sqrt {b x +a}\, a^{2} b^{2} d^{2} x -144 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} c^{2}-160 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} c d x -56 \sqrt {x}\, \sqrt {b x +a}\, a \,b^{3} d^{2} x^{2}+96 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c^{2} x +128 \sqrt {x}\, \sqrt {b x +a}\, b^{4} c d \,x^{2}+48 \sqrt {x}\, \sqrt {b x +a}\, b^{4} d^{2} x^{3}+105 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{4} d^{2}-240 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{3} b c d +144 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) a^{2} b^{2} c^{2}}{192 b^{5}} \] Input:

int(x^2*(d*x+c)^2/(b*x^2+a*x)^(1/2),x)
 

Output:

( - 105*sqrt(x)*sqrt(a + b*x)*a**3*b*d**2 + 240*sqrt(x)*sqrt(a + b*x)*a**2 
*b**2*c*d + 70*sqrt(x)*sqrt(a + b*x)*a**2*b**2*d**2*x - 144*sqrt(x)*sqrt(a 
 + b*x)*a*b**3*c**2 - 160*sqrt(x)*sqrt(a + b*x)*a*b**3*c*d*x - 56*sqrt(x)* 
sqrt(a + b*x)*a*b**3*d**2*x**2 + 96*sqrt(x)*sqrt(a + b*x)*b**4*c**2*x + 12 
8*sqrt(x)*sqrt(a + b*x)*b**4*c*d*x**2 + 48*sqrt(x)*sqrt(a + b*x)*b**4*d**2 
*x**3 + 105*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**4*d* 
*2 - 240*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**3*b*c*d 
 + 144*sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a))*a**2*b**2*c* 
*2)/(192*b**5)